Two coils P and Q are placed near each other. When a current of 3 A flows through coil P (no current in Q), the magnetic flux linked with Q is \( 10^{-3}\,\mathrm{Wb} \). When no current passes through P and a current of 2 A passes through Q, the flux linked with P is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 6.67\times10^{-4}\,\mathrm{Wb} \)
The flux linked with one coil due to the current in the other is proportional to that current, \( \Phi=MI \), where \( M \) is the mutual inductance. The mutual inductance is the same both ways, \( M_{12}=M_{21} \), so the value found from one arrangement holds for the other.
From the current in P and the flux linked with Q:
\( \begin{aligned} M &= \frac{\Phi_Q}{I_P} \\ &= \frac{10^{-3}\,\mathrm{Wb}}{3\,\mathrm{A}} \\ &\approx 3.33\times10^{-4}\,\mathrm{H} \end{aligned} \)
With a current of 2 A in Q, the flux linked with P is:
\( \begin{aligned} \Phi_P &= MI_Q \\ &= \frac{10^{-3}}{3}\,\mathrm{H}\times2\,\mathrm{A} \\ &\approx 6.67\times10^{-4}\,\mathrm{Wb} \end{aligned} \)
NCERT (2026-27):
Mutual Inductance – “… we get
\( M_{12}=M_{21} \)\( =M\ \text{(say)} \qquad (6.12) \)
We have demonstrated this equality for long co-axial solenoids. However, the relation is far more general.”